# Isosceles triangle 9

Given an isosceles triangle ABC where AB= AC. The perimeter is 64cm and altitude is 24cm. Find the area of the isosceles triangle

Result

A =  168 cm2

#### Solution:

$p=64 \ \text{cm} \ \\ h=24 \ \text{cm} \ \\ \ \\ p=2a + b \ \\ \ \\ a^2=(b/2)^2 + h^2 \ \\ a^2=((p-2a)/2)^2 + h^2 \ \\ a^2=(p/2-a)^2 + h^2 \ \\ a^2=p^2/4-pa + a^2 + h^2 \ \\ pa=p^2/4 + h^2 \ \\ \ \\ a=p/4 + h^2/p=64/4 + 24^2/64=25 \ \text{cm} \ \\ \ \\ b=p - 2 \cdot \ a=64 - 2 \cdot \ 25=14 \ \text{cm} \ \\ \ \\ p_{1}=2 \cdot \ a + b=2 \cdot \ 25 + 14=64 \ \text{cm} \ \\ p_{1}=p \ \\ \ \\ A=\dfrac{ b \cdot \ h }{ 2 }=\dfrac{ 14 \cdot \ 24 }{ 2 }=168 \ \text{cm}^2$

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Showing 2 comments:
Steve
No, your solution starts with h=25cm, but the question says that h=24cm.

So, looking at one-half of the isosceles triangle, x2 + 242 = (32-x)2    solves to x=7cm,   which gives the area of the isosceles triangle as 7x24 = 168cm2

Dr Math
thank you, we corrected 25 to 24 as altitude....

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